Question

11. Isaac throws an apple straight up (in the positive direction) from 1.0 m above
the ground, reaching a maximum height of 35 meters. Neglecting air
resistance, what is the ball's velocity when it hits the ground?​

Answer 1

Answer:

26.204 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

At the maximum height initial velocity will be zero.

Equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 35+0^2}\\\Rightarrow v=26.204\ m/s$

If the maximum height is 35 m above the ground then the ball's velocity when it reaches the ground is 26.204 m/s.

Answer 2

Answer:

26.19m/s(down)

Explanation:

distance=35m

acceleration= -9.8m/s^2

vf^2=vi^2 +2ad

vf^2= 0+2(-9.8)(35)

vf^2=-625

vf= $\sqrt{625}$

vf=26.19m/s(down)